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d="M18.8,85.1h56l0,0c2.2,0,4-1.8,4-4v-32h-8v28h-48v-48h28v-8h-32l0,0c-2.2,0-4,1.8-4,4v56C14.8,83.3,16.6,85.1,18.8,85.1z"></path> <polygon fill="currentColor" points="45.7,48.7 51.3,54.3 77.2,28.5 77.2,37.2 85.2,37.2 85.2,14.9 62.8,14.9 62.8,22.9 71.5,22.9"></polygon></svg> <span class="sr-only">(opens new window)</span></span></a></p></blockquote> <blockquote><p><strong>题意：</strong></p> <p>给定两个大小为 m 和 n 的有序数组<code>nums1</code>和<code>nums2</code>。</p> <p>请你找出这两个有序数组的中位数，并且要求算法的时间复杂度为 O(log(m + n))。</p> <p>你可以假设<code>nums1</code>和<code>nums2</code>不会同时为空。</p> <p><strong>示例1：</strong></p> <div class="language- line-numbers-mode"><pre class="language-text"><code>nums1 = [1, 3]
nums2 = [2]

则中位数是 2.0
</code></pre> <div class="line-numbers-wrapper"><span class="line-number">1</span><br><span class="line-number">2</span><br><span class="line-number">3</span><br><span class="line-number">4</span><br></div></div><p><strong>示例2：</strong></p> <div class="language- line-numbers-mode"><pre class="language-text"><code>nums1 = [1, 2]
nums2 = [3, 4]

则中位数是 (2 + 3)/2 = 2.5
</code></pre> <div class="line-numbers-wrapper"><span class="line-number">1</span><br><span class="line-number">2</span><br><span class="line-number">3</span><br><span class="line-number">4</span><br></div></div></blockquote> <h2 id="一、准备"><a href="#一、准备" class="header-anchor">#</a> 一、准备</h2> <p>题目已经给出<strong>有序数组</strong>和算法的<strong>时间复杂度必须为 O(log(m + n))</strong>。</p> <h3 id="时间复杂度"><a href="#时间复杂度" class="header-anchor">#</a> 时间复杂度</h3> <p>O(1)即为<strong>一次查找</strong>即能获取想要的结果，O(n)为要<strong>遍历所有元素</strong>才能获取想要的结果。</p> <p>O(log n)则常出现在<strong>二分查找（也叫折半查找）中</strong>，因为每次查找都是一半长度一半长度的查找，比如查找一个长度为<strong>16</strong>的数组中某一元素，需要<code>4</code>次，<code>(1 / 2) ^ 4 = 1 / 16</code>，<code>n = 16</code>，那么查找次数就是<strong>以2为底，n的对数</strong>-&gt;<code>log n</code>。</p> <p>那么O(log(m + n))中，<code>m + n</code>即为两个有序数组的长度之和，该时间复杂度就意味着需使用<strong>二分/折半</strong>查找。</p> <h2 id="二、解"><a href="#二、解" class="header-anchor">#</a> 二、解</h2> <div class="language-js line-numbers-mode"><pre class="language-js"><code><span class="token comment">/**
 * @param {number[]} nums1
 * @param {number[]} nums2
 * @return {number}
 */</span>
<span class="token keyword">var</span> <span class="token function-variable function">findMedianSortedArrays</span> <span class="token operator">=</span> <span class="token keyword">function</span><span class="token punctuation">(</span><span class="token parameter">nums1<span class="token punctuation">,</span> nums2</span><span class="token punctuation">)</span> <span class="token punctuation">{</span>
    <span class="token keyword">let</span> arr <span class="token operator">=</span> <span class="token punctuation">[</span><span class="token operator">...</span>nums1<span class="token punctuation">,</span> <span class="token operator">...</span>nums2<span class="token punctuation">]</span><span class="token punctuation">.</span><span class="token function">sort</span><span class="token punctuation">(</span><span class="token punctuation">(</span><span class="token parameter">a<span class="token punctuation">,</span> b</span><span class="token punctuation">)</span> <span class="token operator">=&gt;</span> a <span class="token operator">-</span> b<span class="token punctuation">)</span><span class="token punctuation">;</span><span class="token comment">// 将两个数组合并 并 排序</span>
    <span class="token keyword">let</span> length <span class="token operator">=</span> arr<span class="token punctuation">.</span>length<span class="token punctuation">;</span>
  	<span class="token comment">// 如果数组长度为 奇数 取中间的哪一个，为 偶数 则取中间两个求平均</span>
    <span class="token keyword">return</span> length <span class="token operator">%</span> <span class="token number">2</span> <span class="token operator">?</span> arr<span class="token punctuation">[</span>Math<span class="token punctuation">.</span><span class="token function">floor</span><span class="token punctuation">(</span>length <span class="token operator">/</span> <span class="token number">2</span><span class="token punctuation">)</span><span class="token punctuation">]</span> <span class="token operator">:</span> <span class="token punctuation">(</span>arr<span class="token punctuation">[</span>length <span class="token operator">/</span> <span class="token number">2</span><span class="token punctuation">]</span> <span class="token operator">+</span> arr<span class="token punctuation">[</span>length <span class="token operator">/</span> <span class="token number">2</span> <span class="token operator">-</span> <span class="token number">1</span><span class="token punctuation">]</span><span class="token punctuation">)</span> <span class="token operator">/</span> <span class="token number">2</span><span class="token punctuation">;</span>
<span class="token punctuation">}</span><span class="token punctuation">;</span>
</code></pre> <div class="line-numbers-wrapper"><span class="line-number">1</span><br><span class="line-number">2</span><br><span class="line-number">3</span><br><span class="line-number">4</span><br><span class="line-number">5</span><br><span class="line-number">6</span><br><span class="line-number">7</span><br><span class="line-number">8</span><br><span class="line-number">9</span><br><span class="line-number">10</span><br><span class="line-number">11</span><br></div></div><p><strong>提交结果：</strong></p> <table><thead><tr><th>Time</th> <th>Cache</th></tr></thead> <tbody><tr><td>144ms</td> <td>39.8MB</td></tr></tbody></table></div> <footer class="page-edit"><!----> <div class="last-updated"><span class="prefix">Last Updated:</span> <span class="time">2 years ago</span></div></footer> <div class="page-nav"><p class="inner"><span class="prev"><a href="/blog/leetcode/js/exercises-05.html" class="prev"><i aria-label="图标: left" class="anticon anticon-left"><svg viewBox="64 64 896 896" focusable="false" data-icon="left" width="1em" height="1em" fill="currentColor" aria-hidden="true"><path d="M724 218.3V141c0-6.7-7.7-10.4-12.9-6.3L260.3 486.8a31.86 31.86 0 0 0 0 50.3l450.8 352.1c5.3 4.1 12.9.4 12.9-6.3v-77.3c0-4.9-2.3-9.6-6.1-12.6l-360-281 360-281.1c3.8-3 6.1-7.7 6.1-12.6z"></path></svg></i>
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